Problem: Simplify and expand the following expression: $ \dfrac{1}{2q + 6}+ \dfrac{1}{q + 6}+ \dfrac{4q}{q^2 + 9q + 18} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $2$ out of denominator in the first term: $ \dfrac{1}{2q + 6} = \dfrac{1}{2(q + 3)}$ We can factor the quadratic in the third term: $ \dfrac{4q}{q^2 + 9q + 18} = \dfrac{4q}{(q + 3)(q + 6)}$ Now we have: $ \dfrac{1}{2(q + 3)}+ \dfrac{1}{q + 6}+ \dfrac{4q}{(q + 3)(q + 6)} $ The least common multiple of the denominators is: $ 2(q + 3)(q + 6)$ In order to get the first term over $2(q + 3)(q + 6)$ , multiply by $\dfrac{q + 6}{q + 6}$ $ \dfrac{1}{2(q + 3)} \times \dfrac{q + 6}{q + 6} = \dfrac{q + 6}{2(q + 3)(q + 6)} $ In order to get the second term over $2(q + 3)(q + 6)$ , multiply by $\dfrac{2(q + 3)}{2(q + 3)}$ $ \dfrac{1}{q + 6} \times \dfrac{2(q + 3)}{2(q + 3)} = \dfrac{2(q + 3)}{2(q + 3)(q + 6)} $ In order to get the third term over $2(q + 3)(q + 6)$ , multiply by $\dfrac{2}{2}$ $ \dfrac{4q}{(q + 3)(q + 6)} \times \dfrac{2}{2} = \dfrac{8q}{2(q + 3)(q + 6)} $ Now we have: $ \dfrac{q + 6}{2(q + 3)(q + 6)} + \dfrac{2(q + 3)}{2(q + 3)(q + 6)} + \dfrac{8q}{2(q + 3)(q + 6)} $ $ = \dfrac{ q + 6 + 2(q + 3) + 8q} {2(q + 3)(q + 6)} $ Expand: $ = \dfrac{q + 6 + 2q + 6 + 8q}{2q^2 + 18q + 36} $ $ = \dfrac{11q + 12}{2q^2 + 18q + 36}$